Boundedness

Upper Bound

Given

S \subseteq P

where

P

is a partial order, then

u \in P

is said to be an upper bound iff

\forall s \in S, s \leq u

. we say that

S

is bounded above by

u

Lower Bound

Given

S \subseteq P

where

P

is a partial order , then

l \in P

is said to be an lower bound iff

\forall s \in S, l \leq s

, in this case we say that

S

is bounded below by

l

Bounded

A set

S \subseteq P

where

P

is a partial order, then

P

is said to be bounded if it is bounded above and bounded below

Bounded Function

we say that

f : X \to R

is bounded on

S \subseteq X

when

f (S)

is bounded

Infimum

The infimum of a subset

S

of a partially ordered set

P

is the greatest element in

P

that is a lower bound of

S

and is denoted by

inf (S)

. Symbolically this means that

\forall s \in S, s \geq inf (S)

, and if

k

is a lower bound, then

k \leq inf (S)

Supremum

The supremum of a subset

S

of a partially ordered set

P

is the least element in

P

that is a upper bound of

S

and is denoted by

sup (S)

. Symbolically this means that

\forall s \in S, s \leq sup (S)

, and if

k

is a upper bound, then

k \geq sup (S)

Supremum of Suprema Is the Same as Their Union

Suppose

P

is a partially ordered set and that

T \subseteq 𝒫 (P)

and that for each

S \in T, \sup (S)

exists in

P

, then if

\sup (⋃ T) exists and equals x ⟺ \sup ({\sup (S) : S \in T}) exists and equals x

We may also denote the infimum as the GLB (greatest lower bound) and the supremum as the LUB (least upper bound), note that not all sets have a LUB or a GLB.

Supremum is Unique

As per title.

The supremum is the smallest element of a set, therefore it is unique.

Greater than Infimum Implies Not a Lower Bound

Suppose that

inf (S) < x

, then

x

is not an lower bound of

S

Suppose that $inf (S) < x$ , then for the sake of contradiction assume that $x$ is a lower bound of $S$ , then $inf (S) \geq x$ which is a contradiction.

less than sup implies not an upper bound

x < sup (S)

, then

x

is not a upper bound of

S

Suppose that $sup (S) > x$ , then for the sake of contradiction assume that $x$ is a upper bound of $S$ , then $sup (S) \leq x$ which is a contradiction.

Infimum or Supremum of a Set may not be part of Itself

Given a set

S

it's possible that

inf (S) \notin S

sup (S) \notin S

Consider the set

S = (0, 1) \subseteq ℝ

, then

sup (S) = 1, inf (S) = 0

and

1 \notin S, 0 \notin S

Maximum Value

Given a set

S

a maximum value is an element

s \in S

such that for every

s^{'} \in S, s^{'} \leq s

One way of thinking of the maximum value is that the set $S$ contains one of it's upper bounds.

Least Upper Bound Property

We say that a partially ordered set

X

has the least upper bound property if every non-empty subset of

X

that has an upper bound, has a least upper bound in

X

, moreover it is unique. Symbolically that is, if

S \subseteq X

and

S \neq \emptyset

, if

S

has an upper bound, then

sup (S) \in S

If an Upper Bound is Contained in the Set it is the Supremum

As per title.

Suppose we have some upper bound

s \in S

such that for every

s^{'} \in S

s^{'} \leq s

, then

s

is the supremum of

S

, the reason why is that by contradiction if there was a smaller upper bound say

x

such that

x < s

then

x

would not be an upper bound because there exists an element in

S

greater than it, namely

s

, therefore

s

is the smallest upper bound and so

\sup (S) = s

Maximum Value iff Contains Sup

A set

S

has a maximum value given by

\sup (S)

iff

\sup (S) \in S

$⟹$ Suppose $S$ has a maximum value $s$ therefore it is the supremum, therefore since the maximum value is contained within the set $S$ then s

$⟸$ Let $s = \sup (S)$ and assume that $s \in S$ therefore $s$ is an upper bound and is contained in $S$ , therefore $s$ is the max value of $S$ as needed.

A Number is Greater than the Other iff the Difference is Positive

Suppose that

a, b \in ℝ

then

a < b ⟺ 0 < b - a

Multiplying an Inequality by Minus One Flips it

Suppose that

a, b \in ℝ

then

a < b ⟺ - a > - b

a < b

is equivalent to

0 < b - a = (- a) - (- b)

which is equivalent to

- a > - b

, as needed.

Negative of a Set

Suppose that

S \subseteq ℝ

then we define

- S = {- s : s \in S}

Negative of Infimum is Supremum of Negatives

Let

\emptyset \neq S \subseteq ℝ

, be bounded below, then

- \inf (S) = \sup (- S)

Define $α = \inf (S)$ and $β = \sup (- S)$ , by definition $α$ is the greatest lower bound of $S$ this means that for every $s \in S$ we know that $α \leq s$ which is to say that $- a \geq - s$ which shows that $- α$ is an upper bound to the set $- S$ therefore $- α \geq β$ as $β$ is the least upper bound.

Symmetrically $β$ is the least upper bound of $- S$ which means that $β \geq - s$ for every $s \in S$ thus we have $- β \leq s$ so that $- β$ is a lower bound of $S$ since $α$ is the greatest lower bound then we have that $- β \leq α$ so then $β \geq - α$

Thus from the above two paragraphs we conclude that $β = - α$ in other words $- \inf (S) = \sup (- S)$

If a Set Equals its Negative Then its Absolute Value is Its Non-Negative Half

Suppose that

X \subseteq ℝ

and that

X = - X

then

| X | = X \cap ℝ^{\geq 0}

Let $a \in | X |$ therefore $a = | x |$ for some $x \in X$ if $x < 0$ then $| x | = - x$ , since $- X = X$ then we know that $- x \in X$ so that $a = | x | = - x \in X$ as needed.

Now assume that $a \in X \cap ℝ^{\geq 0}$ , therefore $a \in X$ and $a \geq 0$ since $a \in X$ then $a = | a | \in | X |$ as needed.

If a Set Equals its Negative the Sup of the Absolute value and the Original Set are the Same

Suppose that

X \subseteq ℝ

and that

X = - X

then

\sup (| X |) = \sup (X)

X = {0}

then this holds true trivially, otherwise

X

has at least one positive element or at least one negative element, suppose that it has a positive element

p \in X

then since

\sup (X)

is an upper bound of

X

we know that

\sup (X) > 0

and so we know that

\sup (X) = \sup (X \cap ℝ^{\geq 0})

, at the same time we already knew that

| X | = X \cap ℝ^{\geq 0}

so all in all we have

\sup (X) = \sup (X \cap ℝ^{\geq 0}) = \sup (| X |)

Now if

X

didn't have a positive element, but instead it had a negative element, then we perform the above analysis on the set

- X

Half Length Interval

Let

a, b \in ℝ

such that

a < b

, and set

M = \frac{a + b}{2}

, then

ℓ ([a, M]) = ℓ ([M, b]) = \frac{ℓ ([a, b])}{2}

Recall that

ℓ ([a, b]) = b - a

and now see that

ℓ ([a, M]) = M - a = \frac{a + b}{2} - a = \frac{b - a}{2} = \frac{ℓ ([a, b])}{2}

similarly

ℓ ([M, b]) = b - M = b - \frac{a + b}{2} = \frac{b - a}{2} = \frac{ℓ ([a, b])}{2}

ℝ

has the Least Upper Bound Property

ℝ

has the least upper bound property

Let $S \subseteq ℝ$ and $S \neq \emptyset$ , because of that let $s \in S$ and set $a_{0} = s$ , quickly note that if it so happens that $a_{0}$ is an upper bound then we can take $a_{0}$ to be our least upper bound. We also know that $S$ is bounded above, so let $b_{0}$ be any upper bound of $S$ , consider $d : = b_{0} - a_{0}$ , and note that we must have $d > 0$ , for if not then $b_{0} < a_{0}$ which would mean that $b_{0}$ is not an upper bound. If $d = 0$ then $a_{0} = b_{0}$ , and so as before $S$ has the supremum $a_{0}$ .

Therefore we can assume that $a_{0} < b_{0}$ and also assume that $a_{0}$ is not an upper bound. Now consider the value $M_{n} : = \frac{a_{n} + b_{n}}{2}$ . Let's define two sequences simultaneously, if $M_{n}$ is an upper bound of $S$ then define $a_{n + 1} = a_{n}$ and $b_{n + 1} = M_{n}$ , if $M_{n}$ is not an upper bound of $S$ then we define $a_{n + 1} = M_{n}$ and $b_{n + 1} = b_{n}$ . The goal of this is to consider $M_{n}$ as an approximation of our supremum, if its too big we adjust our upper bound to force the next iteration to get closer, if it's inside the set we push up our lower bound to force it outside the set.

We claim that every $n \in ℕ_{0}, b_{n}$ is an upper bound of $S$ and that every $a_{n}$ is not an upper bound of $S$ . We prove both at once through induction, we assumed $a_{0}$ was not an upper bound, and $b_{0}$ by definition was so the base case holds. Assume that the statement holds true for $k \in ℕ_{0}$ and we'll show that it holds true for $k + 1$ . Suppose that $M_{k}$ is an upper bound to $S$ , in that case $b_{k + 1} = M_{k}$ and so $b_{k + 1}$ is an upper bound, in this case $a_{k + 1} = a_{k}$ and since we knew that $a_{k}$ was not an upper bound $a_{k + 1}$ is also not an upper bound, in the case that $M_{k}$ is not an upper bound, then $b_{k + 1} = b_{k}$ and so $b_{k + 1}$ is still an upper bound, but since $a_{k + 1} = M_{k}$ we see that $a_{k + 1}$ still remains not an upper bound as needed.

Consider the sequence of intervals $I_{n} : = [a_{n}, b_{n}]$ and define $d_{j} = ℓ (I_{j})$ then we claim that $d_{j} = \frac{d_{0}}{2^{j}}$ For $j = 0$ it holds trivially, then suppose for $k \in ℕ_{0}$ it holds true, and we'll prove it holds true for $j + 1$ , so we know that $ℓ ([a_{j}, b_{j}]) = \frac{d_{0}}{2_{j}}$ . If $M_{j}$ is an upper bound then $a_{j + 1} = a_{j}$ and $b_{j + 1} = M_{j}$ therefore $ℓ ([a_{j + 1}, b_{j + 1}]) = \frac{d_{0}}{2_{j}} \cdot (\frac{1}{2}) = \frac{d_{0}}{2^{j + 1}}$ as needed, the case when $M_{j}$ is not an upper bound follows symmetrically, thus the statement holds true for all $j \in ℕ_{0}$ .

By the definition of $a_{n}, b_{n}$ it follows that for every $n, m \in ℕ_{0}$ we have that $a_{n} < b_{m}$ for if there was a $k, j \in ℕ_{0}$ such that $a_{k} \geq b_{j}$ since $b_{j}$ is an upper bound, so is $a_{k}$ , but that's a contradiction because we proved that $a_{k}$ is not an upper bound for any $k$ . Also consider through the definition of $a_{n}, b_{n}$ that we must have that $a_{n} \leq a_{n + 1}$ and that $b_{n} \leq b_{n + 1}$ . This shows that for any $J \in ℕ_{0}$ and any $j \geq J, a_{j}, b_{j} \in [a_{J}, b_{J}]$ because we know that $a_{J} \leq a_{j} \leq b_{j} \leq b_{J}$

Now consider the sequence $(c_{n}) = (a_{0}, b_{0}, a_{1}, b_{1}, a_{2}, b_{2}, \dots)$ , defined so that

$c_{n} = a_{\frac{n}{2}}$ if $n$ even
$c_{n} = b_{\frac{n}{2} - 1}$ if $n$ odd

We will show that this sequence is cauchy. So let $ϵ \in ℝ^{+}$ and let $K \in ℕ_{0}$ such that $d 2^{- K} < ϵ$ , note that we know that $ℓ (I_{K}) = \frac{d_{0}}{2^{K}}$ , take $N = 2 K + 2$ and let $n, m \geq N$ we must prove that $| c_{n} - c_{m} | \leq ϵ$ , since we know that $\frac{d_{0}}{2^{K}} \leq ϵ$ it's enough to show that $c_{n}, c_{m} \in I_{K}$ , recall that $c_{n}$ is either equal to $c_{\frac{n}{2}}$ or $b_{\frac{n}{2} - 1}$ but since $N \geq 2 K + 2$ we see that the indices are greater than $K$ that is $\frac{n}{2} > \frac{n}{2} - 1 \geq \frac{2 K + 2}{2} - 1 = K$ therefore no matter what $c_{n}, c_{m} \in I_{K}$ as hinted at earlier, in other words $| c_{n} - c_{m} | \leq \frac{d_{0}}{2^{K}}$ as needed, so that $(c_{n})$ is cauchy.

Since it is cauchy it converges to some limit $L$ , we will prove that $L$ is the supremum of $S$ , that is, it is the greatest upper bound, first of all it must be an upper bound, for if it wasn't then there would be some $s \in S$ such that $s > L$ , recall that since $(c_{n}) \to L$ then $(b_{n}) \to L$ as it's a subsequence, in that case consider $ϵ = \frac{s - L}{2}$ , then there exists some $N \in ℕ_{0}$ such that $| b_{N} - L | < ϵ$ so that $b_{N} < s$ but this implies that $b_{N}$ is not an upper bound which is a contradiction, therefore $L$ must be an upper bound.

But it also is the least upper bound, so suppose for the sake of contradiction that there was a smaller upper bound $B$ so that $B < L$ and set $ϵ = \frac{L - B}{2}$ similarly to our previous paragraph we can obtain some $N$ such that $| a_{N} - L | < ϵ$ which implies that $B < a_{N}$ but that would imply that $a_{N}$ is an upper bound, which is not possible, and therefore we have a contradiction so $L$ is the least upper bound.